What is the integral from t to 0 of 1/(1+x^2) ?

The integral from t to 0 of 1/(1+x^2) is -arctan(t)

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integral from t to 0 of 1/(1+x^2)

\int_{t}^{0} \frac{1}{1 + x ^{2}} d x

- arctan (t)

Solution steps

\int_{t}^{0} \frac{1}{1 + x ^{2}} d x

Use the common integral:

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

= [arctan (x)]_{t}^{0}

Compute the boundaries:

- arctan (t)

= - arctan (t)

f (x) = \frac{40}{0.01 x ^{2} + 1}

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What is the integral from t to 0 of 1/(1+x^2) ?
The integral from t to 0 of 1/(1+x^2) is -arctan(t)